>Subject: Re: Conditional Probability Question>From: Virgil vmhjr2@home.com >Date: 9/25/01 11:23 PM Central Daylight Time>Message-id: <vmhjr2-DECB08.22331625092001@news1.denver1.co.home.com>>>In article <20010925234846.05795.00000711@mb-mf.aol.com>,> mensanator@aol.com (Mensanator) wrote:>>> >Subject: Re: Conditional Probability Question>> >From: Ray Vickson rvickson@engmail.uwaterloo.ca >> >Date: 9/25/01 5:09 PM Central Daylight Time>> >Message-id: <3BB10091.36487FAB@engmail.uwaterloo.ca>>> >>> >There is a difference between examining previous parts and not examining>> >them. If you take out (i-1) parts and put them aside without examining>> >them, the probability that the i'th part (which IS examined) will be>> >defective is 1/100. Surprisingly, this is true even if i=100, so you are>> >drawing from a box that contains only 1 part. However, everything>> >changes if the set-aside parts are examined instead of ignored.>> >> Why does it change?>>One's evaluation of probability is generally dependent on one's >state of knowledge at the time of the evaluation. Each part >examined, up to finding the defective part, changes one's state of >knowledge.

I don't see that. What changes is that on every inspection, the probabilty offinding the part goes from 1/100 to 1/1. But that is _not_ the probability thatthe part is _at_ position i. It is the probability of finding it at postion i _combined_ with the probability that it is _not_ at posititions i-1. And if youdo the correct math, you'll see that this is 1/100 for every i.

And the probability the ith part is defective does not change based on yourknowledge of the previous i-1 inspections, so it doesn't matter whether youinspect them or not.